JEE Main 2026 — 6 April, Evening Shift — Question 45

$\mathrm{N}_{\text {atoms }}$ in $\mathrm{O}_{2}=\frac{\mathrm{m}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{O}_{2}}} \mathrm{~N}_{\mathrm{A}}=\frac{2}{32} \times 2 \mathrm{~N}_{\mathrm{A}}=\frac{\mathrm{N}_{\mathrm{A}}}{8}=0.125 \mathrm{~N}_{\mathrm{A}}$ $\mathrm{N}_{\text {atoms }}$ in $\mathrm{SO}_{2}=\frac{\mathrm{m}_{\mathrm{SO}_{2}}}{\mathrm{M}_{\mathrm{SO}_{2}}}=3 \mathrm{~N}_{\mathrm{A}}=\frac{4}{64} \times 3 \mathrm{~N}_{\mathrm{A}}=\frac{3 \mathrm{~N}_{\mathrm{A}}}{16}=0.1875 \mathrm{~N}_{\mathrm{A}}$ $\mathrm{N}_{\text {atoms }}$ in $\mathrm{O}_{2}$ at STP $=\frac{\mathrm{V}_{\mathrm{O}_{2}}}{22.7 \text { lit }} \mathrm{N}_{\mathrm{A}}=\frac{1400 \mathrm{ml}}{22400 \mathrm{ml}} \times 2 \mathrm{~N}_{\mathrm{A}}=0.125 \mathrm{~N}_{\mathrm{A}}$ $\mathrm{N}_{\text {atoms }}$ in $\mathrm{He}(\mathrm{g})$ at $\mathrm{STP}=\frac{\mathrm{V}_{\mathrm{He}}}{22.4 \text { lit. }} \mathrm{N}_{\mathrm{A}}=\frac{0.05}{22.4} \mathrm{~N}_{\mathrm{A}}=0.002 \mathrm{~N}_{\mathrm{A}}$ $\mathrm{N}_{\text {atoms }}$ in $\mathrm{H}_{2}(\mathrm{~g})=$ moles × atomicity $\times \mathrm{N}_{\mathrm{A}}=0.0625 \times 2 \mathrm{~N}_{\mathrm{A}}$